Question 1026253: $20,862 is invested, part at 9% and the rest at 7%. If the interest earned from the amount invested at 9% exceeds the interest earned from the amount invested at 7% by $831.50, how much is invested at each rate?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! There is an amount invested at 9%. We don't know how much it is, so we will call it x.
There is another amount invested at 7%, and we know that this is equal to 20,862-x.
And the problem says:
0.09x = 0.07(20,862-x)+831.50 Multiply on right to get rid of parenthesis
0.09x = 1,460.34-0.07x+831.50
0.09x = 2,291.84-0.07x add 0.07x on both sides:
0.16x = 2,291.84 divide both sides by 0.16
x = 14,324 this is the amount invested at 9%
And:
20,862-14,324 = 6,538 is invested at 7%
Happy learning,
J
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